Year 8 Edexcel Statistics: Unit Test Mock Paper Walkthrough | 八年级爱德思统计：单元测试模拟卷解析

📚 Year 8 Edexcel Statistics: Unit Test Mock Paper Walkthrough | 八年级爱德思统计：单元测试模拟卷解析

This article provides a detailed walkthrough of a typical Year 8 Edexcel Statistics unit test mock paper. Each section tackles a key topic, presenting a model question followed by clear, bilingual explanations to reinforce understanding and exam technique.

本文详细解析一份典型的八年级爱德思统计单元测试模拟卷。每个小节围绕一个核心考点，给出典型题目并配以清晰的中英双语解答，旨在巩固知识并强化应试技能。

1. Designing a Questionnaire | 设计问卷

Question: You want to find out how Year 8 students spend their free time after school. Write two suitable questions for a questionnaire, each with a choice of at least three response boxes. Explain why your questions avoid bias.

题目：你想了解八年级学生课后如何度过闲暇时间。为问卷设计两个合适的问题，每个问题提供至少三个选项框，并解释你的问题为何避免了偏差。

Good question 1: “On a typical school day, how many hours do you spend on leisure activities (e.g. reading, gaming, sports)? 0–1 hour, 1–2 hours, 2–3 hours, more than 3 hours.”

好问题1：“在通常的上学日，你花在休闲活动（如阅读、游戏、运动）上的时间是多少？0–1小时，1–2小时，2–3小时，超过3小时。”

Good question 2: “Which of these activities do you enjoy most after school? Reading / Sports / Screen time / Creative hobbies (e.g. music, art).”

好问题2：“放学后你最喜欢下列哪项活动？阅读 / 运动 / 屏幕时间 / 创造性爱好（如音乐、美术）。”

Why these avoid bias: The response options are specific, mutually exclusive and cover a range of possibilities without leading the respondent towards a particular answer. The wording is neutral and does not imply one activity is better than another.

为何避免偏差：选项具体、相互排斥且涵盖多种可能性，不会把回答者引向某个特定答案。措辞中性，不暗示某项活动优于其他。

2. Bar Charts and Frequency | 条形图与频数

Example: The table below shows the favourite colours of 45 students. Use the data to draw a bar chart. Which colour is the mode?

例题：下表显示了45名学生最喜爱的颜色。用数据画出条形图。哪种颜色是众数？

Colour	Frequency
Red	12
Blue	18
Green	10
Yellow	5

Step 1: Label the horizontal axis with the colour categories and the vertical axis with frequency, scaling it up to at least 18.

步骤1：横轴标上颜色类别，纵轴标上频数，刻度至少到18。

Step 2: Draw bars of equal width for each colour. The height of each bar must match its frequency: Red 12, Blue 18, Green 10, Yellow 5.

步骤2：为每种颜色画等宽的直条。每一条的高度必须对应频数：红12，蓝18，绿10，黄5。

Step 3: Add a title, e.g. “Favourite colours of Year 8 students”. The bar for Blue is the tallest, so the mode is Blue.

步骤3：添加标题，例如“八年级学生最喜爱的颜色”。蓝条最高，因此众数是蓝色。

3. Pie Charts and Angles | 饼图与角度

Question: 30 students were asked about their pets. The results are: Dog 12, Cat 9, Fish 6, No pet 3. Calculate the angle for each sector and draw the pie chart.

题目：30名学生接受了宠物调查。结果：狗12人，猫9人，鱼6人，无宠物3人。计算每个扇形的角度并画出饼图。

Total frequency = 12 + 9 + 6 + 3 = 30. One student represents 360° ÷ 30 = 12°.

总频数 = 12 + 9 + 6 + 3 = 30。每名学生代表 360° ÷ 30 = 12°。

Dog angle = 12 × 12° = 144°. Cat angle = 9 × 12° = 108°. Fish angle = 6 × 12° = 72°. No pet angle = 3 × 12° = 36°.

狗扇区角度 = 12 × 12° = 144°。猫扇区 = 9 × 12° = 108°。鱼扇区 = 6 × 12° = 72°。无宠物扇区 = 3 × 12° = 36°。

Check: 144° + 108° + 72° + 36° = 360°. Draw the circle, measure each angle with a protractor, label each sector and add a title.

检验：144° + 108° + 72° + 36° = 360°。画出圆，用量角器量出各角度，标出每个扇区并加上标题。

4. Stem-and-Leaf Diagrams | 茎叶图

Data: 23, 25, 28, 31, 31, 34, 36, 40, 42. Draw an ordered stem-and-leaf diagram and find the median.

数据：23, 25, 28, 31, 31, 34, 36, 40, 42。画出有序茎叶图并找出中位数。

Step 1: Use the tens digit as the stem and units digit as the leaf. Stem 2: leaves 3, 5, 8. Stem 3: leaves 1, 1, 4, 6. Stem 4: leaves 0, 2. Always order the leaves from smallest to largest.

步骤1：十位数字作茎，个位数字作叶。茎2：叶3, 5, 8。茎3：叶1, 1, 4, 6。茎4：叶0, 2。务必把叶从小到大排序。

Step 2: Include a key, e.g. “2 | 3 means 23”. The ordered diagram makes it easy to find the median. There are 9 values, so the median is the 5th value: 31.

步骤2：添加图例，例如“2 | 3 表示 23”。有序茎叶图便于寻找中位数。共有9个数值，中位数是第5个：31。

5. Scatter Graphs and Correlation | 散点图与相关性

Question: The table shows hours spent revising and test scores for 5 students. Plot the points on a scatter graph. Describe the type of correlation. Predict the score for a student who revises for 7 hours.

题目：下表记录了5名学生的复习时间与测试成绩。在散点图上描点。描述相关性的类型。预测复习7小时的学生的成绩。

Revision (hours)	2	3	4	5	6
Test score (%)	50	55	65	70	75

Plot each pair (hours, score) as a cross. The points slope upwards, showing a positive correlation: as revision hours increase, test score tends to increase.

将每对数据（复习时间，成绩）用叉号画出。各点呈上升趋势，呈正相关：复习时间增加，测试成绩往往也提高。

To predict a score for 7 hours, we can extend the trend line. Following the pattern, a score of roughly 80–85% would be a sensible estimate.

要预测7小时的成绩，可沿趋势线延伸。依据模式，约80–85%是合理的估计值。

6. Mean, Median, Mode and Range | 平均数、中位数、众数和极差

Find the mean, median, mode and range of this data set: 12, 15, 20, 22, 22, 25, 30.

求下列数据集的平均数、中位数、众数和极差：12, 15, 20, 22, 22, 25, 30。

Mean: Sum = 12+15+20+22+22+25+30 = 146. Number of values = 7. Mean = 146 ÷ 7 ≈ 20.9 (to one decimal place).

平均数：总和 = 12+15+20+22+22+25+30 = 146。数据个数 = 7。平均数 = 146 ÷ 7 ≈ 20.9（保留一位小数）。

Median: Ordered list, 4th value is 22, so median = 22. Mode: 22 appears twice, all others once, so mode = 22. Range: 30 – 12 = 18.

中位数：有序列表中第4个值为22，所以中位数 = 22。众数：22出现两次，其它均一次，所以众数 = 22。极差：30 – 12 = 18。

7. Probability Scale | 概率尺度

Mark the approximate probability of each event on a probability line labelled 0, 1/2 and 1: a) Flipping a fair coin and getting heads; b) Drawing a heart from a standard 52-card deck; c) The sun rising tomorrow morning.

在标有0、1/2和1的概率线上，标出每个事件的大致概率：a) 抛一枚公平硬币得到正面；b) 从标准52张牌中抽到红心；c) 明天早晨太阳升起。

Event a: P(heads) = 1/2, so place mark exactly at the midpoint. Event b: There are 13 hearts, so P(heart) = 13/52 = 1/4, which is closer to 0 than to 1/2. Mark it one quarter of the way from 0. Event c: The sun rising is virtually certain, P ≈ 1, so mark at the far right end.

事件a：P(正面) = 1/2，标记刚好在中点。事件b：共有13张红心，P(红心) = 13/52 = 1/4，比1/2更接近0，标记在从0起四分之一处。事件c：太阳升起几乎必然，P ≈ 1，标记在最右端。

8. Sample Space Diagrams | 样本空间图

Two fair spinners are spun. Spinner A has numbers 1, 2, 3, 4; Spinner B has numbers 1, 2, 3, 4. List all possible outcomes in a sample space diagram. Find the probability that the sum of the two numbers is 5.

转动两个公平的转盘。转盘A标有1、2、3、4；转盘B标有1、2、3、4。用样本空间图列出所有可能结果。求两数之和为5的概率。

There are 4 × 4 = 16 equally likely outcomes. Outcomes with sum of 5 are: (1,4), (2,3), (3,2), (4,1) — four favourable outcomes.

共有 4 × 4 = 16 种等可能结果。和为5的结果有：(1,4), (2,3), (3,2), (4,1) — 四个有利结果。

Therefore, P(sum = 5) = 4/16 = 1/4. The sample space diagram helps verify that no outcomes are missed.

因此，P(和为5) = 4/16 = 1/4。样本空间图有助于确保不遗漏任何结果。

9. Two-Way Tables | 双向表

80 students are asked which sport they prefer: football or basketball. 24 boys prefer football, 14 boys prefer basketball. 16 girls prefer football. Complete the two-way table and find the probability that a randomly chosen student is a girl who prefers football.

80名学生被问及喜欢足球还是篮球。24名男生喜欢足球，14名男生喜欢篮球。16名女生喜欢足球。完成双向表并求随机选到的学生是喜欢足球的女生的概率。

Boys total = 24 + 14 = 38. Total students = 80, so total girls = 80 – 38 = 42. Girls who prefer basketball = 42 – 16 = 26.

男生总数 = 24 + 14 = 38。学生总数 = 80，则女生总数 = 80 – 38 = 42。喜欢篮球的女生 = 42 – 16 = 26。

The completed table: Football (Boys 24, Girls 16, Total 40); Basketball (Boys 14, Girls 26, Total 40). P(girl and football) = 16/80 = 1/5.

完成后的表格：足球（男生24，女生16，合计40）；篮球（男生14，女生26，合计40）。P(喜欢足球的女生) = 16/80 = 1/5。

10. Interpreting Statistical Graphs | 统计图的解读

The line graph below shows the average monthly temperature in two cities, A and B, over a year. Use the graph to answer: In which month is the difference in temperature between the cities greatest? Compare the temperature trends.

下面的折线图显示A、B两城市一年的月平均气温。看图回答：哪个月两城市温差最大？比较气温变化趋势。

Examine the vertical gap between the two lines each month. The greatest gap appears in August, where City A records around 28°C and City B around 18°C, a difference of roughly 10°C.

逐月观察两条折线的垂直间隔。最大差距出现在八月，A城市约28°C，B城市约18°C，相差约10°C。

Trend: City A has a clear summer peak from June to August and colder winters. City B shows a more moderate, steady temperature throughout the year with a smaller range. Both cities reach their highest temperatures around July.

趋势：A城市6至8月有明显的夏季高峰，冬季较冷。B城市全年气温较为温和平稳，温度范围较小。两城市最高温均出现在七月前后。

11. Mixed Exam-Style Question | 综合考试风格题

A fair six-sided die is rolled 30 times. The frequency of each score is shown: 1 (3 times), 2 (7 times), 3 (5 times), 4 (6 times), 5 (4 times), 6 (5 times). Calculate the relative frequency of rolling an odd number. Is the die likely to be fair? Explain.

一枚公平六面骰子掷了30次。各点频数如下：1（3次）、2（7次）、3（5次）、4（6次）、5（4次）、6（5次）。计算掷出奇数的相对频率。该骰子是否可能公平？解释。

Odd scores are 1, 3, 5. Total odd frequency = 3 + 5 + 4 = 12. Relative frequency = 12/30 = 2/5 = 0.4.

奇数是1、3、5。奇数总频数 = 3 + 5 + 4 = 12。相对频率 = 12/30 = 2/5 = 0.4。

For a fair die, we would expect P(odd) = 1/2 = 0.5. The experimental relative frequency of 0.4 is somewhat lower, but with only 30 trials, some variation is normal. More trials would be needed to confidently conclude bias.

对公平骰子，我们期望P(奇数) = 1/2 = 0.5。实验相对频率0.4略低，但仅30次试验存在波动是正常的。需要更多试验才能确信骰子有偏。

Published by TutorHao | Statistics Revision Series | aleveler.com

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